It is well known (see the section 14 of [1]}) that the subgroup of \(\displaystyle \mathrm {E}_6\)  generated by \(\displaystyle \mathrm {SL}_3\)  and \(\displaystyle \mathrm {Spin_8}\)  is \(\displaystyle \mathrm {E}_6\)  itself. In fact, one can prove that the subgroup of \(\displaystyle \mathrm {E}_6\)  generated by \(\displaystyle \mathrm {SL}_3\)  and \(\displaystyle \mathrm {Spin_7}\)  is also \(\displaystyle \mathrm {E}_6\) . The following argument has been communicated to me by Robert Bryant. Let \(\displaystyle h \in \mathrm {SL}_3\)  defined by:
\( h = \begin{pmatrix} 0 & -1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & -1 \end{pmatrix}.\) 
